Mechanics of a falling slinky

So I saw a video of a falling slinky in slow motion where a slinky is held at the top and is let to be stretched by gravity until gravity and spring pulling force equal out, when the top is let free, the bottom stays still until the top actually falls down reaching the bottom and discharging its tension and only then the bottom (with the top) start to fall down ..
Pretty interesting stuff.
I have a few questions about this.
*When the top part finally reaches the bottom part, does it accelerate the bottom part, apart from gravity, with the kinetic energy it gained when the spring collapses? On that note - does upper part fall down with just the acceleration of gravity or gravity + spring pulling?

When the top part finally reaches the bottom part, does it accelerate the bottom part, apart from gravity, with the kinetic energy it gained when the spring collapses?

The upper turns impact on the ones below. However, since after the upper turns impact, they fall down close together, the impact is quite inelastic. This means that a lot of energy transforms into heat and not to kinetic energy of the moving parts.

On that note - does upper part fall down with just the acceleration of gravity or gravity + spring pulling?

The spring is stretched, so the gravity is assisted by string in pulling the upper parts.

The acceleration of the bottom part of the slinky will be due to the sum of all forces acting on it. So there would definitely be gravity, but also whatever force the rest of the slinky is imparting to it with tension/compression.

The instant after the top is let go, the only force acting on the slinky as a whole is gravity, which causes its center of gravity to accelerate down at 9.8 m/s/s. The center of mass acceleration will be the same at every point the slinky is falling until the bottom hits the ground, but the slinky is also pulling itself together as it is no longer under tension. Thus, the top of the slinky will be falling with a larger acceleration, and the bottom part of the slinky will be falling more slowly.

If the bottom of the slinky is stationary before the rest of the slinky catches up to it, the bottom part of the slinky will accelerate quickly to the same speed as the rest of the falling slinky, an acceleration due to gravity in addition to the force of the rest of the slinky coming into contact with it.

Ok, that is a very good answer.
*What if I put a mass at the bottom of the slinky (and the slinky stretches further down to a new point of equilibrium) and, just to make it more interesting, mold the mass from the attachment point at the bottom all around the hanging slinky in a form of a box. (lets assume the slinky is held from within the box by a "ghost" hand that can actually enter the closed box and hold it) - will the whole box experience this .. levitation for a split second?

The answer to that is probably "yes" ..

What I really want to understand is - if I stretch and mold the hanging mass around the slinky just so it barely fits the slinky inside the box (or cylinder) . In fact, the top part of the slinky would be a the same height as the top part (the roof) of the box. Not just that, but the slinky would be attached to the roof of the box.
Now, instead of a ghost hand holding it, we can assume the slinky was stretched to the proper length (to the length the slinky would be stretched by gravity if it was hanging with the given mass attached) on the ground by scientists, enclosed in this box and the box lifted in the air by, say, a rocket.. and the slinky is released from the attachment to the roof of the box at the same time as the rocket stops pushing up.

Now what? The slinky will contract and try to pull the bottom part with the whole attached mass up, but the bottom would stay in place because gravity is pulling down at the same force. The bottom with the whole surrounding box.. ?

umm, a link to a pdf .. well, thank you, but would be really nice if I get a actual response about this exact thing? Like a discussion, you know.

Well, sorry for that, I actually started typing a longish reply about how a spring is described by a wave equation, which has a general solution of a waveform traveling up and down the spring, how the boundary conditions at the ends cause waveform to be reflected back, how the shape of the waveform is determined by the initial conditions and is in fact quadratic with the sharp discontinuity, how this causes the bottom end to stay put until the wave propagates all the way from the top and then sharply move down, how a slinky is different from a spring etc. etc.
...but then accidentally lost everything I typed and I though well surely someone must have already written all this up so with the right kind of search keywords the first hit on google gave me this article which pretty much sums up everything I was going to write in the first place.

I mean this is a well-known solved problem, what is there to discuss. If you cannot follow the math, sorry, you would just have to read the rest of the article and take the conclusions for granted. If you don't know what wave equation is, or why it is applicable to the case at hand, look it up on wikipedia, it actually gives explicit derivation for the case of a massive spring.
If you have specific questions please ask.

....
Now what? The slinky will contract and try to pull the bottom part with the whole attached mass up, but the bottom would stay in place because gravity is pulling down at the same force. The bottom with the whole surrounding box.. ?

The answer is yes. The string is initially in equilibrium. When the top of the spring is released, the disturbance travels down at the speed c=l sqrt(k/m) according to the wave equation. Until it reaches the bottom at the time t = sqrt(m/k), the bottom of the spring (and the attached mass) remains in equilibrium and does not accelerate. The shape of the attached mass (whether it surrounds the spring or simply hangs below it) does not matter.

The shape of the spring at the moment of release is important and that depends on the motion of the rocket before the release. If the rocket travels at a constant speed (or hovers) long enough for the spring to settle into equilibrium, then the answer is yes, the box will continue moving at the same speed (or hovering) for some brief time after release.

OR you can, while on the ground, suspend the box from the spring, wait for it to settle and then attach the spring to the box not only at the top but along the whole length of the spring. Then you throw the whole thing up in the air and at the top of trajectory when the velocity of the box is 0, you release the spring, leaving it attached at the bottom only.